function plusD(n) {
if (n > 9) {
let varies = 0;
let NewN = n.toString().split('');
console.log(NewN)
for (let i = 0; i < NewN.length; i++) {
varies = varies + parseInt(NewN[i])
}
if (varies > 9) {
plusD(varies)
}
return varies;
}
}
As you can see, in the end there is an extra if statement used , when true calls the plusD function again which results in another addition of the number or if false(only when the number has been successfully reduced to single digit), returns the number.
You are missing a return. Without that, the function return value will never be taken into account.
function plusD(n) {
if (n > 9) {
let varies = 0;
let NewN = n.toString().split('');
console.log(NewN)
for (let i = 0; i < NewN.length; i++) {
varies = varies + parseInt(NewN[i])
}
if (varies > 9) {
return plusD(varies)
}
return varies;
}
}
console.log(plusD(29));
function plusD(n) {
var varies;
if (n > 9) {
varies = 0;
let NewN = n.toString().split('');
for (let i = 0; i < NewN.length; i++) {
varies = varies + parseInt(NewN[i])
}
if (varies > 9) {
return plusD(varies)
}
}
return varies;
}
console.log(plusD(123456));
Related
Trying my hardest to improve but it's been almost 3 hours and I still haven't figured out why this function returns a bunch of 0's
I'm suppose to check the hirsch number of a given array. It is the greatest number n in which n is n times greater than or equal to the values of the array.
This is what I have so far
// Hirsch Number given an array with each space in the array represents a paper and each number represents the amount of citations for a given paper.
function hirschNumber(numberArray)
{
let hirschIndex = 0;
let continueCheck = true;
for (i=0; i < numberArray.length; i++)
{
let count = 0;
let citationsTreshold = hirschIndex + 1
for (i=0; i < numberArray.length; i++)
{
if (citationsTreshold <= numberArray[i])
{
count++
}
if (numberArray.length == (i - 1))
{
if (count >= citationsTreshold)
{
hirschIndex++
}
else if (count < citationsTreshold)
{
continueCheck = false
return
}
}
if (continueCheck = false)
{
return hirschIndex
}
else
{
console.log(hirschIndex)
}
}
if (continueCheck = false)
{
return hirschIndex
}
else {
console.log(hirschIndex)
};
};
return hirschIndex;
};
console.log(hirschNumber([3,0,6,1,5]));
I have an issue with a recursive algorithm, that solves the problem of finding the happy numbers.
Here is the code:
function TestingFunction(number){
sumNumberContainer = new Array(0);
CheckIfNumberIsHappy(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
console.log(sumOfTheNumbers);
if(sumOfTheNumbers == 1){
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
//return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
//return false;
}
}
}
}
CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
Algorithm is working ALMOST fine. I've tested it out by calling function with different numbers, and console was displaying correct results. The problem is that I almost can't get any value from the function. There are only few cases in which I can get any value: If the number is build out of ,,0", and ,,1", for example 1000.
Because of that, I figured out, that I have problem with returning any value when the function is calling itself again.
Now I ended up with 2 results:
Returning the
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
which is giving an infinity looped number. For example when the number is happy, the function is printing in the console number one again and again...
Returning the
//return true
or
//return false
which gives me an undefined value
I'm a little bit in check by this problem, and I'm begging you guys for help.
I would take a step back and reexamine your problem with recursion in mind. The first thing you should think about with recursion is your edge cases — when can you just return a value without recursing. For happy numbers, that's the easy case where the sum of squares === 1 and the harder case where there's a cycle. So test for those and return appropriately. Only after that do you need to recurse. It can then be pretty simple:
function sumSq(num) {
/* simple helper for sums of squares */
return num.toString().split('').reduce((a, c) => c * c + a, 0)
}
function isHappy(n, seen = []) {
/* seen array keeps track of previous values so we can detect cycle */
let ss = sumSq(n)
// two edge cases -- just return
if (ss === 1) return true
if (seen.includes(ss)) return false
// not an edge case, save the value to seen, and recurse.
seen.push(ss)
return isHappy(ss, seen)
}
console.log(isHappy(23))
console.log(isHappy(22))
console.log(isHappy(7839))
Here's a simplified approach to the problem
const digits = x =>
x < 10
? [ x ]
: [ ...digits (x / 10 >> 0), x % 10 ]
const sumSquares = xs =>
xs.reduce ((acc, x) => acc + x * x, 0)
const isHappy = (x, seen = new Set) =>
x === 1
? true
: seen.has (x)
? false
: isHappy ( sumSquares (digits (x))
, seen.add (x)
)
for (let n = 1; n < 100; n = n + 1)
if (isHappy (n))
console.log ("happy", n)
// happy 1
// happy 7
// happy 10
// ...
// happy 97
The program above could be improved by using a technique called memoization
Your code is almost correct. You just forgot to return the result of the recursive call:
function TestingFunction(number){
sumNumberContainer = new Array(0);
if (CheckIfNumberIsHappy(number))
console.log(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
if(sumOfTheNumbers == 1){
return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return false;
}
}
}
}
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
for (let i=0; i<100; ++i)
TestingFunction(i.toString()); // 1 7 10 13 ... 91 94 97
I've got the solution, which was given to me in the comments, by the user: Mark_M.
I just had to use my previous
return true / return false
also I had to return the recursive statement in the function, and return the value of the CheckIfTheNumberIsHappy function, which was called in TestingFunction.
The working code:
function TestingFunction(number){
sumNumberContainer = new Array(0);
return CheckIfNumberIsHappy(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
console.log(sumOfTheNumbers);
if(sumOfTheNumbers == 1){
return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return false;
}
}
}
}
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
Thanks for the great support :)
I wrote a simple script for a website called Codewars (here: https://www.codewars.com/kata/57814d79a56c88e3e0000786). The purpose of the function was to encrypt a string such that every second character would appear first, and then the rest of them. I tested many random strings of text; it worked for a while. But then, I tested a specific case with 17 characters: "maybe do i really", and it resulted in a character being dropped (notably a space). Initially, I thought the issue was that the .join method didn't allow a double space in a row, so I attempted to make my own function to mimic its functionality: it did not solve the problem. Could anyone answer why this specific string loses a character and returns a wrong encryption? My jsfiddle is here: https://jsfiddle.net/MCBlastoise/fwz62j2g/
Edit: I neglected to mention that it runs a certain number of times based on parameter n, encrypting the string multiple times per that value.
And my code is here:
function encrypt(text, n) {
if (n <= 0 || isNaN(n) === true || text === "" || text === null) {
return text;
}
else {
for (i = 1; i <= n; i++) {
if (i > 1) {
text = encryptedString;
}
var evenChars = [];
var oddChars = [];
for (j = 0; j < text.length; j++) {
if (j % 2 === 0) {
evenChars.push(text.charAt(j));
}
else {
oddChars.push(text.charAt(j));
}
}
var encryptedString = oddChars.join("") + evenChars.join("");
}
return encryptedString;
}
}
function decrypt(encryptedText, n) {
if (n <= 0 || encryptedText === "" || encryptedText === null) {
return encryptedText;
}
else {
for (i = 1; i <= n; i++) {
if (i > 1) {
encryptedText = decryptedString;
}
var oddChars = [];
var evenChars = [];
for (j = 0; j < encryptedText.length; j++) {
if (j < Math.floor(encryptedText.length / 2)) {
oddChars.push(encryptedText.charAt(j));
}
else {
evenChars.push(encryptedText.charAt(j));
}
}
var convertedChars = []
for (k = 0; k < evenChars.length; k++) {
convertedChars.push(evenChars[k]);
convertedChars.push(oddChars[k]);
}
var decryptedString = convertedChars.join("");
}
return decryptedString;
}
}
document.getElementById("text").innerHTML = encrypt("maybe do i really", 1);
document.getElementById("text2").innerHTML = decrypt("ab oiralmyed ely", 1)
<p id="text"></p>
<p id="text2"></p>
Nothing wrong with the code itself. Basically HTML doesn't allow 2 or more spaces. You can use <pre> tag for the case like this.
document.getElementById("text").innerHTML = "<pre>" + encrypt("maybe do i really", 1) + "</pre>";
I'm trying to make a factor calculator. You input a number, and it finds out the factors of that number. If you divide the original number by its factor you get zero, and I'm trying to implement that here so that when it returns with '0' it gets pushed to an array and that array is printed.
var number = prompt("Number?")
var array = []
function modulo(a, b)
{
return a % b;
}
for (counter = 0; counter < number; counter++)
{
var result = modulo(number, counter)
if (result = 0)
{
array.push(counter)
}
}
for (counter = 0; counter < array.length; counter++)
{
alert(array[counter])
}
What happens is the prompt shows up, I input a number, and nothing happens. Can anybody help?
Here is the problem, you have used = (Assignment operator) instead of == comparison operator
for (counter = 0; counter < number; counter++)
{
var result = modulo(number, counter)
if (result == 0) // in your code this is result = 0
{
array.push(counter)
}
}
Working demo
Complete code:
var number = prompt("Number?")
var array = []
function modulo(a, b)
{
return a % b
}
for (counter = 0; counter < number; counter++)
{
var result = modulo(number, counter)
if (result == 0)
{
array.push(counter)
}
}
for (counter = 0; counter < array.length; counter++)
{
alert(array[counter])
}
To check if values are equal, use == and not =:
if (result == 0)
I'm creating a form where users can input a range. They are allowed to input letters and numbers. Some sample input:
From: AA01
To: AZ02
Which should result in:
AA01
AA02
AB01
AB02
And so on, till AZ02
And:
From: BC01
To: DE01
Should result in:
BC01
BD01
BE01
CC01
CD01
CE01
Etc
I managed to get it working for the input A01 to D10 (for example)
jsFiddle
However, i can't get it to work with multiple letters.
JS code:
var $from = $('input[name="from"]');
var $to = $('input[name="to"]');
var $quantity = $('input[name="quantity"]');
var $rangeList = $('.rangeList');
var $leadingzeros = $('input[name="leadingzeros"]');
$from.on('keyup blur', function () {
$(this).val($(this).val().replace(/[^a-zA-Z0-9]/g, ''));
updateQuantity();
});
$to.on('keyup blur', function () {
$(this).val($(this).val().replace(/[^a-zA-Z0-9]/g, ''));
updateQuantity();
});
$leadingzeros.on('click', function () {
updateQuantity();
});
function updateQuantity() {
var x = parseInt($from.val().match(/\d+/));
var y = parseInt($to.val().match(/\d+/));
var xl = $from.val().match(/[a-zA-Z]+/);
var yl = $to.val().match(/[a-zA-Z]+/);
var result = new Array();
if (xl != null && yl != null && xl[0].length > 0 && yl[0].length > 0) {
xl = xl[0].toUpperCase();
yl = yl[0].toUpperCase();
$rangeList.html('');
var a = yl.charCodeAt(0) - xl.charCodeAt(0);
for (var i = 0; i <= a; i++) {
if (!isNaN(x) && !isNaN(y)) {
if (x <= y) {
var z = (y - x) + 1;
$quantity.val(z * (a + 1));
$rangeList.html('');
for (var b = z; b > 0; b--) {
var c = ((y - b) + 1);
if ($leadingzeros.prop('checked')) {
c = leadingZeroes(c, y.toString().length);
}
result.push(String.fromCharCode(65 + i) + c);
}
} else {
$rangeList.html('');
$quantity.val(0);
}
} else {
$rangeList.html('');
$quantity.val(0);
}
}
} else if (!isNaN(x) && !isNaN(y)) {
if (x < y) {
var z = (y - x) + 1;
$quantity.val(z);
$rangeList.html('');
for (var i = z; i > 0; i--) {
var c = (y - i) + 1;
if ($leadingzeros.prop('checked')) {
c = leadingZeroes(c, y.toString().length);
}
result.push(c);
}
} else {
$rangeList.html('');
$quantity.val(0);
}
} else {
$rangeList.html('');
$quantity.val(0);
}
$rangeList.html('');
for (var i = 0; i < result.length; i++) {
$rangeList.append(result[i] + '<br />');
}
}
function leadingZeroes(number, size) {
number = number.toString();
while (number.length < size) number = "0" + number;
return number;
}
This is perfect for a recursive algorithm:
function createRange(from, to) {
if (from.length === 0) {
return [ "" ];
}
var result = [];
var innerRange = createRange(from.substring(1), to.substring(1));
for (var i = from.charCodeAt(0); i <= to.charCodeAt(0); i++) {
for (var j = 0; j < innerRange.length; j++) {
result.push(String.fromCharCode(i) + innerRange[j]);
}
}
return result;
}
Called as follows:
createRange('BC01', 'DE02'); // Generates an array containing all values expected
EDIT: Amended function below to match new test case (much more messy, however, involving lots of type coercion between strings and integers).
function prefixZeroes(value, digits) {
var result = '';
value = value.toString();
for (var i = 0; i < digits - value.length; i++) {
result += '0';
}
return result + value;
}
function createRange(from, to) {
if (from.length === 0) {
return [ "" ];
}
var result = [];
if (from.charCodeAt(0) < 65) {
fromInt = parseInt(from);
toInt = parseInt(to);
length = toInt.toString().length;
var innerRange = createRange(from.substring(length), to.substring(length));
for (var i = fromInt; i <= toInt; i++) {
for (var j = 0; j < innerRange.length; j++) {
result.push(prefixZeroes(i, length) + innerRange[j]);
}
}
} else {
var innerRange = createRange(from.substring(1), to.substring(1));
for (var i = from.charCodeAt(0); i <= to.charCodeAt(0); i++) {
for (var j = 0; j < innerRange.length; j++) {
result.push(String.fromCharCode(i) + innerRange[j]);
}
}
}
return result;
}
Please note that because of your strict logic in how the value increments this method requires exactly 4 characters (2 letters followed by 2 numbers) to work. Also, this might not be as efficient/tidy as it can be but it took some tinkering to meet your logic requirements.
function generate(start, end) {
var results = [];
//break out the start/end letters/numbers so that we can increment them seperately
var startLetters = start[0] + start[1];
var endLetters = end[0] + end[1];
var startNumber = Number(start[2] + start[3]);
var endNumber = Number(end[2] + end[3]);
//store the start letter/number so we no which value to reset the counter to when a maximum boundry in reached
var resetLetter = startLetters[1];
var resetNumber = startNumber;
//add first result as we will always have at least one
results.push(startLetters + (startNumber < 10 ? "0" + startNumber : "" + startNumber));
//maximum while loops for saefty, increase if needed
var whileSafety = 10000;
while (true) {
//safety check to ensure while loop doesn't go infinite
whileSafety--;
if (whileSafety == 0) break;
//check if we have reached the maximum value, if so stop the loop (break)
if (startNumber == endNumber && startLetters == endLetters) break;
//check if we have reached the maximum number. If so, and the letters limit is not reached
//then reset the number and increment the letters by 1
if (startNumber == endNumber && startLetters != endLetters) {
//reset the number counter
startNumber = resetNumber;
//if the second letter is at the limit then reset it and increment the first letter,
//otherwise increment the second letter and continue
if (startLetters[1] == endLetters[1]) {
startLetters = '' + String.fromCharCode(startLetters.charCodeAt(0) + 1) + resetLetter;
} else {
startLetters = startLetters[0] + String.fromCharCode(startLetters.charCodeAt(1) + 1);
}
} else {
//number limit not reached so just increment the number counter
startNumber++;
}
//add the next sequential value to the array
results.push(startLetters + (startNumber < 10 ? "0" + startNumber : "" + startNumber));
}
return results;
}
var results = generate("BC01", "DE01");
console.log(results);
Here is a working example, which uses your second test case
Using #Phylogenesis' code, i managed to achieve my goal.
jsFiddle demo
function updateQuantity() {
var x = parseInt($from.val().match(/\d+/));
var y = parseInt($to.val().match(/\d+/));
var xl = $from.val().match(/[a-zA-Z]+/);
var yl = $to.val().match(/[a-zA-Z]+/);
var result = new Array();
var r = createRange(xl[0], yl[0]);
var z = (y - x) + 1;
if (x <= y) {
for (var j = 0; j < r.length; j++) {
var letters = r[j];
for (var i = z; i > 0; i--) {
var c = (y - i) + 1;
if ($leadingzeros.prop('checked')) {
c = leadingZeroes(c, y.toString().length);
}
if (i == z) {
r[j] = letters + c + '<br />';
} else {
j++;
r.splice(j, 0, letters + c + '<br />');
}
}
}
} else {
for (var i = 0; i < r.length; i++) {
r[i] += '<br />';
}
}
$quantity.val(r.length);
$rangeList.html('');
for (var i = 0; i < r.length; i++) {
$rangeList.append(r[i]);
}
}
This works for unlimited letters and numbers, as long as the letters are first.
Thanks for your help!